Well, kind of.

The point is that infinite energies correspond to unphysical degrees of freedom. This is what the integral above says---for example, if we are computing a simple process like electron + positron goes to muon + anti-muon. One can compute quantum corrections to the classical result, and end up with integrals like the one above.

The quantum corrections are loop diagrams: for example, electron plus positron can anihilate into a photon, which can turn into a muon and anti-muon, like this:

The corrections are quantum mechanical effects that aren't taken care of in the classical limit. They look like this, for example:

The quantum process here is the creation of a ``virtual'' electron positron pair in the middle---that is, the photon which is being exchanged turns into an electron and positron, which quickly anihilate back into the photon. They are called ``virtual'' because we cannot see them, but we know that they're there---we can calculate and measure their effects.

When we compute the effects of the virtual electron positron pair, we are left with integrals that look like

$$\int_0^{\infty} dp \rightarrow \infty$$

which we encoutered in the previous post. The problem is that these virtual processes DO occur, but the answers that we measure are finite. That is, this process contributes to something called Compton scattering, which we CAN measure very accurately.

Does this mean that all of QFT is wrong?

Of course not. The resolution is the fact that we are integrating over unphysical momenta---the integral goes all the way up to infinity, but there is clearly no way for the electron/positron in the loop to have infinite momentum.

The main point of all of this is that you can think of renormalization as a procedure to cut off an unphysical region of your integral. There are much more fancy procedures for doing this, but in general you could do something like this:

$$\int_0^{\infty} dp \rightarrow \int_0^{M} dp $$,

where M is called the cutoff. The amazing thing you find is that the answer

*doesn't depend on M*!!! So what you have done, is you have cut off a region of parameter space which is unphysical, and your final answer doesn't depend on how you did the cutoff in the first place.

This is why I don't understand Hawking's comments---he said (aparently) that ``Renormalization is not proven''. But there's nothing to prove---all that you're doing is limiting the theory to a place where you expect it to apply. For example, the Bohr model example that I used before. You wouldn't think of trying to compute the electron energies of uranium with such a model---it JUST doesn't apply. Similarly, you wouldn't think of computing QED with infinite photon energies---it just doesn't apply there.